Question

Using vector, prove that $\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta$

Answer 1

Let $P\left(\alpha \right)$ and $Q(\frac{\pi }{2}+\beta )$ are trigonometric points
$\therefore P\left(\alpha \right)=(\cos \alpha ,\sin \alpha )$
And $Q(\frac{\pi }{2}+\beta )=\left(\cos (\frac{\pi }{2}+\beta )\right),\sin (\frac{\pi }{2}+\beta )=(-\sin \beta ,\cos \beta )$
$\overrightarrow{OP}=(\cos \alpha ,\sin \alpha )$
$\overrightarrow{OQ}=(-\sin \beta ,\cos \beta )$
Angle between $\overrightarrow{OP}.\overrightarrow{OQ}=\frac{\pi }{2}+\beta -\alpha =\frac{\pi }{2}-(\alpha -\beta )$
$\therefore \cos [\frac{\pi }{2}-(\alpha -\beta )]=\frac{\overrightarrow{OP}.\overrightarrow{OQ}}{\left|\overrightarrow{OP}\right|\left|\overrightarrow{OQ}\right|}$
$\therefore \sin (\alpha -\beta )=\frac{(\cos \alpha ,\sin \alpha ).(-\sin \beta ,\cos \beta )}{\left(\sqrt{{\cos }^2\alpha +{\sin }^2\beta }\right)\sqrt{{\sin }^2\beta +{\cos }^2\beta }}$
$\therefore \sin (\alpha -\beta )=-\cos \alpha \sin \beta +\sin \alpha \cos \beta$
$\therefore \sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta$